Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{2y - 18}{-2y + 18} \times \dfrac{-3y^3 + 15y^2 + 108y}{y^3 + 5y^2 + 4y} $
Explanation: First factor out any common factors. $n = \dfrac{2(y - 9)}{-2(y - 9)} \times \dfrac{-3y(y^2 - 5y - 36)}{y(y^2 + 5y + 4)} $ Then factor the quadratic expressions. $n = \dfrac {2(y - 9)} {-2(y - 9)} \times \dfrac {-3y(y + 4)(y - 9)} {y(y + 4)(y + 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {2(y - 9) \times -3y(y + 4)(y - 9) } {-2(y - 9) \times y(y + 4)(y + 1) } $ $n = \dfrac {-6y(y + 4)(y - 9)(y - 9)} {-2y(y + 4)(y + 1)(y - 9)} $ Notice that $(y + 4)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-6y\cancel{(y + 4)}(y - 9)(y - 9)} {-2y\cancel{(y + 4)}(y + 1)(y - 9)} $ We are dividing by $y + 4$ , so $y + 4 \neq 0$ Therefore, $y \neq -4$ $n = \dfrac {-6y\cancel{(y + 4)}\cancel{(y - 9)}(y - 9)} {-2y\cancel{(y + 4)}(y + 1)\cancel{(y - 9)}} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $n = \dfrac {-6y(y - 9)} {-2y(y + 1)} $ $ n = \dfrac{3(y - 9)}{y + 1}; y \neq -4; y \neq 9 $